How do you implicitly differentiate #-y= 4x^3y^2+2x^2y^3-2xy^4 #?

Answer 1

#y'=(-12x^2y^3-4xy^3+2y^4)/(1+8x^3y+6x^2y^2-8xy^3)#

Implicit Differentiation is a special case of the chain rule. When you differentiate the y variable in a term or factor you differentiating with respect to x. Because of this you have include the factor of #dy/dx# or #y'#.

We will have to use the product rule, power rule and chain rule.

Divide all the terms by -1 to remove the negative

#y=-4x^3y^2-2x^2y^3+2xy^4#

Differentiate Implicitly

#y'=-4x^3 2yy'-12x^2y^3-(2x^2 3y^2 y'+4xy^3)+2x4y^3y'+2y^4#

Distribute the negative

#y'=-4x^3 2yy'-12x^2y^3-2x^2 3y^2 y'-4xy^3+2x4y^3y'+2y^4#

Gather the terms with the #y'# factors

#y'+4x^3 2yy'+2x^2 3y^2 y'-2x4y^3y'=-12x^2y^3-4xy^3+2y^4#

Factor out #y'#

#y'(1+4x^3 2y+2x^2 3y^2-2x4y^3)=-12x^2y^3-4xy^3+2y^4#

Divide to isolate #y'#

#(y'cancel((1+4x^3 2y+2x^2 3y^2-2x4y^3)))/cancel((1+4x^3 2y+2x^2 3y^2-2x4y^3))=(-12x^2y^3-4xy^3+2y^4)/(1+4x^3 2y+2x^2 3y^2-2x4y^3)#

#y'=(-12x^2y^3-4xy^3+2y^4)/(1+4x^3 2y+2x^2 3y^2-2x4y^3)#

Multiple numeric factors to simplify

#y'=(-12x^2y^3-4xy^3+2y^4)/(1+8x^3y+6x^2y^2-8xy^3)#

Below are a couple of tutorials include implicit differentiation

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Answer 2

To implicitly differentiate the given equation, follow these steps:

  1. Differentiate each term with respect to ( x ).
  2. Apply the product rule when differentiating terms that involve both ( x ) and ( y ).

Differentiating term by term:

For ( -y ), the derivative is ( -\frac{dy}{dx} ).

For ( 4x^3y^2 ), the derivative is ( 12x^2y^2 + 8x^3y\frac{dy}{dx} ) using the product rule.

For ( 2x^2y^3 ), the derivative is ( 4x^2y^3 + 6x^2y^2\frac{dy}{dx} ) using the product rule.

For ( -2xy^4 ), the derivative is ( -2y^4 - 8xy^3\frac{dy}{dx} ) using the product rule.

Combine all the terms:

[ -\frac{dy}{dx} = 12x^2y^2 + 8x^3y\frac{dy}{dx} + 4x^2y^3 + 6x^2y^2\frac{dy}{dx} - 2y^4 - 8xy^3\frac{dy}{dx} ]

Group terms involving ( \frac{dy}{dx} ) together:

[ -\frac{dy}{dx} - 8x^3y\frac{dy}{dx} + 6x^2y^2\frac{dy}{dx} - 8xy^3\frac{dy}{dx} = 12x^2y^2 + 4x^2y^3 - 2y^4 ]

Factor out ( \frac{dy}{dx} ):

[ \frac{dy}{dx}(-1 - 8x^3 + 6x^2y^2 - 8xy^3) = 12x^2y^2 + 4x^2y^3 - 2y^4 ]

Divide both sides by the expression in parentheses:

[ \frac{dy}{dx} = \frac{12x^2y^2 + 4x^2y^3 - 2y^4}{-1 - 8x^3 + 6x^2y^2 - 8xy^3} ]

This is the implicit derivative of the given equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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